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range of values

by sanju09 » Tue Nov 10, 2009 1:51 am
For what range of values of 'x' will the inequality 15 x - (2/x) > 1?
A. x > 0.4
B. x < (1/3)
C. (-1/3)< x < 0.4, x >(15/2)
D. (-1/3)< x < 0.4, x >(25)
E. x < (-1/3) and x > (2/5)
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Re: range of values

by life is a test » Tue Nov 10, 2009 6:59 am
sanju09 wrote:For what range of values of 'x' will the inequality 15 x - (2/x) > 1?
A. x > 0.4
B. x < (1/3)
C. (-1/3)< x < 0.4, x >(15/2)
D. (-1/3)< x < 0.4, x >(25)
E. x < (-1/3) and x > (2/5)
IMO E

15x-2/x > 1 -> 15x^2 - 2 > x -> 15x^2 - x - 2 > x -> (3x+1) (5x-2) -> x > 2/5 or x <-1/3
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by antondesh » Wed Nov 11, 2009 11:15 am
I think it's E as well.
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Re: range of values

by Stuart@KaplanGMAT » Wed Nov 11, 2009 2:12 pm
sanju09 wrote:For what range of values of 'x' will the inequality 15 x - (2/x) > 1?
A. x > 0.4
B. x < (1/3)
C. (-1/3)< x < 0.4, x >(15/2)
D. (-1/3)< x < 0.4, x >(25)
E. x < (-1/3) and x > (2/5)
Picking numbers shows that E can't be correct.

If we let x = -10, we get:

15(-10) - (2/-10) > 1
-150 + 1/5 > 1

which is clearly not true.

Based on the above, we can also eliminate B.

So, let's look at A, C and D:

A. x > 0.4
C. (-1/3)< x < 0.4, x >(15/2)
D. (-1/3)< x < 0.4, x >(25)

Let's think about 0.4, since that's involved in every choice.

If x = 2/5, then we have:

15(2/5) - 2/(2/5) > 1
6 - 10/2 > 1
6 - 5 > 1
1 > 1

So, if x=2/5, we're right "on the post". Therefore, we need to make x either a tiny bit bigger or smaller.

If we increase the value of x, both terms get bigger; if we decrease the value of x (but keep it positive), both terms get smaller. Therefore, we need a value of x greater than .4.

So, x > .4 needs to be part of our solution.

Only A includes that inequality, therefore A must be correct.

* * *

That said, if -1/3 < x < 0, the inequality will also hold true.

If we let x = -1/3, we get:

15(-1/3) - 2/(-1/3) < 1
-5 + 6 < 1
1 < 1

So -1/3 is also a "post".

If we decrease -1/3, both terms become smaller; if we increase -1/3, but keep it negative, both terms become bigger.

So, if the question wanted the full possible range of values for x, the answer should have been:

F. -1/3 < x < 0 or x > 2/5
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Re: range of values

by palvarez » Wed Nov 11, 2009 2:24 pm
sanju09 wrote:For what range of values of 'x' will the inequality 15 x - (2/x) > 1?
A. x > 0.4
B. x < (1/3)
C. (-1/3)< x < 0.4, x >(15/2)
D. (-1/3)< x < 0.4, x >(25)
E. x < (-1/3) and x > (2/5)
x(15x^2 - x -2) > 0
x(15x^2 - 6x + 5x -2) > 0
x[5x(3x+1)-2(3x+1)] > 0
(3x+1)x(5x-2) > 0

The above is +ve in the following sets.

(-1/3, 0) U (2/5, +inf)

A is the answer.
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Re: range of values

by palvarez » Wed Nov 11, 2009 2:30 pm
life is a test wrote:
sanju09 wrote:For what range of values of 'x' will the inequality 15 x - (2/x) > 1?
A. x > 0.4
B. x < (1/3)
C. (-1/3)< x < 0.4, x >(15/2)
D. (-1/3)< x < 0.4, x >(25)
E. x < (-1/3) and x > (2/5)
IMO E

15x-2/x > 1 -> 15x^2 - 2 > x -> 15x^2 - x - 2 > x -> (3x+1) (5x-2) -> x > 2/5 or x <-1/3
you missed one important aspect.

The transformation (like cross multiplication) holds true when x > 0.

When you exclude x > 0 from your answer, you are left with x > 2/5.
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